Problem: The coefficients of the polynomial
\[x^4 + bx^3 + cx^2 + dx + e = 0\]are all integers.  Let $n$ be the exact number of integer roots of the polynomial, counting multiplicity.  For example, the polynomial $(x + 3)^2 (x^2 + 4x + 11) = 0$ has two integer roots counting multiplicity, because the root $-3$ is counted twice.

Enter all possible values of $n,$ separated by commas.
Explanation: The polynomial $x^4 + 1 = 0$ shows that $n$ can be 0

The polynomial $x(x^3 + 2)$ shows that $n$ can be 1.

The polynomial $x^2 (x^2 + 1)$ shows that $n$ can be 2.

The polynomial $x^4$ shows that $n$ can be 4.

Suppose the polynomial has three integer roots.  By Vieta's formulas, the sum of the roots is $-b,$ which is an integer.  Therefore, the fourth root is also an integer, so it is impossible to have exactly three integer roots.

Thus, the possible values of $n$ are $\boxed{0, 1, 2, 4}.$